Find the critical point(s) of the function.

Then use the second derivative test to classify the nature of each point, if possible.
Finally, determine the relative extrema of the function.
f(x,y)=3x^2 - 3xy + 3y^2 +5

To find the critical points of the function f(x, y) = 3x^2 - 3xy + 3y^2 + 5, we need to take the partial derivatives with respect to both x and y and set them equal to zero.

Step 1: Find the partial derivative with respect to x.

∂f/∂x = 6x - 3y

Step 2: Find the partial derivative with respect to y.

∂f/∂y = -3x + 6y

Step 3: Set both partial derivatives equal to zero and solve for x and y.

6x - 3y = 0 -> 2x - y = 0 (Equation 1)
-3x + 6y = 0 -> -x + 2y = 0 (Equation 2)

By solving these two equations simultaneously, we find that x = 0 and y = 0. Therefore, there is one critical point at (0, 0).

Next, let's use the second derivative test to classify the nature of this critical point.

Step 4: Find the second partial derivatives.

∂^2f/∂x^2 = 6
∂^2f/∂y^2 = 6
∂^2f/∂x∂y = -3

Step 5: Substitute the critical point (0, 0) into the second partial derivatives.

∂^2f/∂x^2 (0, 0) = 6
∂^2f/∂y^2 (0, 0) = 6
∂^2f/∂x∂y (0, 0) = -3

Step 6: Calculate the discriminant.

D = (∂^2f/∂x^2)(∂^2f/∂y^2) - (∂^2f/∂x∂y)^2
= (6)(6) - (-3)^2
= 36 - 9
= 27

Step 7: Determine the nature of the critical point using the discriminant.

If D > 0 and (∂^2f/∂x^2) > 0, then the critical point is a local minimum.
If D > 0 and (∂^2f/∂x^2) < 0, then the critical point is a local maximum.
If D < 0, then the critical point is a saddle point.

In this case, D = 27, which is greater than 0. Since (∂^2f/∂x^2) = 6, which is also greater than 0, we can conclude that the critical point (0, 0) is a local minimum.

Finally, let's determine the relative extrema of the function.

Since we have found that the critical point (0, 0) is a local minimum, there are no other critical points to consider. Therefore, the function f(x, y) = 3x^2 - 3xy + 3y^2 + 5 has a relative minimum at (0, 0).